How many permutations of the letters ABCDEFGHI contain
a) the string CDE?
b) the strings AC and CF?
c) the strings ABC and ECG?
Here is my solution:
a) We treat the mentioned string as one single letter. Thus we have the possible letter A,B,CDE,F,G,H,I.
Then we have 7 letters of which we want to select 7 letters.
N=7
R=7
Permutation: P(7,7)=7!/(7-7)!
=7!=5040
b) A string cannot contain both AC and CF, since the string
can only contain C once and C can not be followed by A and F.
Thus there are 0 string that contain both AC and CF.
c) A string cannot contain both ABC and ECG, since the string
can only contain C once and C can not be followed by A and E.
Thus there are 0 string that contain both ABC and ECG.
Is my solution right? or need any modification?
Best Answer
As saulspatz indicated in the comments, your answers to parts (a) and (c) are correct.
In part (b), I think the source of your confusion is that you are assuming that AC and CF must be separate strings. Since you only have one copy of each letter available, for the strings AC and CF to both appear, they must overlap, meaning the permutation must contain the string ACF. Thus, the number of permutations of the letters ABCDEFGHI is the number of permutations which contain the string ACF. Thus, we have seven objects to permute: ACF, B, D, E, G, H, I. The seven objects can be permuted in $7! = 5040$ ways.